Integrand size = 17, antiderivative size = 94 \[ \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx=-\frac {2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}-\frac {10 x^2}{3 b^2 \sqrt {a x+b x^2}}+\frac {5 \sqrt {a x+b x^2}}{b^3}-\frac {5 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{b^{7/2}} \]
-2/3*x^4/b/(b*x^2+a*x)^(3/2)-5*a*arctanh(x*b^(1/2)/(b*x^2+a*x)^(1/2))/b^(7 /2)-10/3*x^2/b^2/(b*x^2+a*x)^(1/2)+5*(b*x^2+a*x)^(1/2)/b^3
Time = 0.07 (sec) , antiderivative size = 96, normalized size of antiderivative = 1.02 \[ \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {x \left (\sqrt {b} x \left (15 a^2+20 a b x+3 b^2 x^2\right )+30 a \sqrt {x} (a+b x)^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}-\sqrt {a+b x}}\right )\right )}{3 b^{7/2} (x (a+b x))^{3/2}} \]
(x*(Sqrt[b]*x*(15*a^2 + 20*a*b*x + 3*b^2*x^2) + 30*a*Sqrt[x]*(a + b*x)^(3/ 2)*ArcTanh[(Sqrt[b]*Sqrt[x])/(Sqrt[a] - Sqrt[a + b*x])]))/(3*b^(7/2)*(x*(a + b*x))^(3/2))
Time = 0.25 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {1133, 1124, 25, 1160, 1091, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 1133 |
\(\displaystyle \frac {5 \int \frac {x^3}{\left (b x^2+a x\right )^{3/2}}dx}{3 b}-\frac {2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1124 |
\(\displaystyle \frac {5 \left (\frac {\int -\frac {a-b x}{\sqrt {b x^2+a x}}dx}{b^2}+\frac {2 a x}{b^2 \sqrt {a x+b x^2}}\right )}{3 b}-\frac {2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {5 \left (\frac {2 a x}{b^2 \sqrt {a x+b x^2}}-\frac {\int \frac {a-b x}{\sqrt {b x^2+a x}}dx}{b^2}\right )}{3 b}-\frac {2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {5 \left (\frac {2 a x}{b^2 \sqrt {a x+b x^2}}-\frac {\frac {3}{2} a \int \frac {1}{\sqrt {b x^2+a x}}dx-\sqrt {a x+b x^2}}{b^2}\right )}{3 b}-\frac {2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {5 \left (\frac {2 a x}{b^2 \sqrt {a x+b x^2}}-\frac {3 a \int \frac {1}{1-\frac {b x^2}{b x^2+a x}}d\frac {x}{\sqrt {b x^2+a x}}-\sqrt {a x+b x^2}}{b^2}\right )}{3 b}-\frac {2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {5 \left (\frac {2 a x}{b^2 \sqrt {a x+b x^2}}-\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a x+b x^2}}\right )}{\sqrt {b}}-\sqrt {a x+b x^2}}{b^2}\right )}{3 b}-\frac {2 x^4}{3 b \left (a x+b x^2\right )^{3/2}}\) |
(-2*x^4)/(3*b*(a*x + b*x^2)^(3/2)) + (5*((2*a*x)/(b^2*Sqrt[a*x + b*x^2]) - (-Sqrt[a*x + b*x^2] + (3*a*ArcTanh[(Sqrt[b]*x)/Sqrt[a*x + b*x^2]])/Sqrt[b ])/b^2))/(3*b)
3.1.61.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_.)/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(3/2), x _Symbol] :> Simp[-2*e*(2*c*d - b*e)^(m - 2)*((d + e*x)/(c^(m - 1)*Sqrt[a + b*x + c*x^2])), x] + Simp[e^2/c^(m - 1) Int[(1/Sqrt[a + b*x + c*x^2])*Exp andToSum[((2*c*d - b*e)^(m - 1) - c^(m - 1)*(d + e*x)^(m - 1))/(c*d - b*e - c*e*x), x], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e ^2, 0] && IGtQ[m, 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(p + 1))), x] - Simp[e^2*((m + p)/(c*(p + 1))) Int[(d + e*x)^(m - 2)*(a + b*x + c*x ^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e ^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntegerQ[2*p]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Time = 1.91 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.87
method | result | size |
pseudoelliptic | \(-\frac {5 \left (a \sqrt {x \left (b x +a \right )}\, \left (b x +a \right ) \operatorname {arctanh}\left (\frac {\sqrt {x \left (b x +a \right )}}{x \sqrt {b}}\right )-\sqrt {b}\, a^{2} x -\frac {4 x^{2} a \,b^{\frac {3}{2}}}{3}-\frac {b^{\frac {5}{2}} x^{3}}{5}\right )}{\sqrt {x \left (b x +a \right )}\, b^{\frac {7}{2}} \left (b x +a \right )}\) | \(82\) |
risch | \(\frac {x \left (b x +a \right )}{b^{3} \sqrt {x \left (b x +a \right )}}-\frac {5 a \ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{2 b^{\frac {7}{2}}}-\frac {2 a^{2} \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{3 b^{5} \left (x +\frac {a}{b}\right )^{2}}+\frac {14 a \sqrt {b \left (x +\frac {a}{b}\right )^{2}-\left (x +\frac {a}{b}\right ) a}}{3 b^{4} \left (x +\frac {a}{b}\right )}\) | \(131\) |
default | \(\frac {x^{4}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {5 a \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {x^{2}}{b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {a \left (-\frac {x}{2 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {1}{3 b \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}-\frac {a \left (-\frac {2 \left (2 b x +a \right )}{3 a^{2} \left (b \,x^{2}+a x \right )^{\frac {3}{2}}}+\frac {16 b \left (2 b x +a \right )}{3 a^{4} \sqrt {b \,x^{2}+a x}}\right )}{2 b}\right )}{4 b}\right )}{2 b}\right )}{2 b}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a x}}-\frac {a \left (-\frac {1}{b \sqrt {b \,x^{2}+a x}}+\frac {2 b x +a}{a b \sqrt {b \,x^{2}+a x}}\right )}{2 b}+\frac {\ln \left (\frac {\frac {a}{2}+b x}{\sqrt {b}}+\sqrt {b \,x^{2}+a x}\right )}{b^{\frac {3}{2}}}}{b}\right )}{2 b}\) | \(268\) |
-5/(x*(b*x+a))^(1/2)/b^(7/2)*(a*(x*(b*x+a))^(1/2)*(b*x+a)*arctanh((x*(b*x+ a))^(1/2)/x/b^(1/2))-b^(1/2)*a^2*x-4/3*x^2*a*b^(3/2)-1/5*b^(5/2)*x^3)/(b*x +a)
Time = 0.25 (sec) , antiderivative size = 221, normalized size of antiderivative = 2.35 \[ \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx=\left [\frac {15 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt {b} \log \left (2 \, b x + a - 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right ) + 2 \, {\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{6 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac {15 \, {\left (a b^{2} x^{2} + 2 \, a^{2} b x + a^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {b x^{2} + a x} \sqrt {-b}}{b x}\right ) + {\left (3 \, b^{3} x^{2} + 20 \, a b^{2} x + 15 \, a^{2} b\right )} \sqrt {b x^{2} + a x}}{3 \, {\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}\right ] \]
[1/6*(15*(a*b^2*x^2 + 2*a^2*b*x + a^3)*sqrt(b)*log(2*b*x + a - 2*sqrt(b*x^ 2 + a*x)*sqrt(b)) + 2*(3*b^3*x^2 + 20*a*b^2*x + 15*a^2*b)*sqrt(b*x^2 + a*x ))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/3*(15*(a*b^2*x^2 + 2*a^2*b*x + a^3)* sqrt(-b)*arctan(sqrt(b*x^2 + a*x)*sqrt(-b)/(b*x)) + (3*b^3*x^2 + 20*a*b^2* x + 15*a^2*b)*sqrt(b*x^2 + a*x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4)]
\[ \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {x^{5}}{\left (x \left (a + b x\right )\right )^{\frac {5}{2}}}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 162 vs. \(2 (78) = 156\).
Time = 0.19 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.72 \[ \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {5 \, a x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {a x}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b^{2}} - \frac {2 \, x}{\sqrt {b x^{2} + a x} a b} - \frac {1}{\sqrt {b x^{2} + a x} b^{2}}\right )}}{6 \, b} + \frac {x^{4}}{{\left (b x^{2} + a x\right )}^{\frac {3}{2}} b} + \frac {10 \, a x}{3 \, \sqrt {b x^{2} + a x} b^{3}} - \frac {5 \, a \log \left (2 \, b x + a + 2 \, \sqrt {b x^{2} + a x} \sqrt {b}\right )}{2 \, b^{\frac {7}{2}}} + \frac {5 \, \sqrt {b x^{2} + a x}}{3 \, b^{3}} \]
5/6*a*x*(3*x^2/((b*x^2 + a*x)^(3/2)*b) + a*x/((b*x^2 + a*x)^(3/2)*b^2) - 2 *x/(sqrt(b*x^2 + a*x)*a*b) - 1/(sqrt(b*x^2 + a*x)*b^2))/b + x^4/((b*x^2 + a*x)^(3/2)*b) + 10/3*a*x/(sqrt(b*x^2 + a*x)*b^3) - 5/2*a*log(2*b*x + a + 2 *sqrt(b*x^2 + a*x)*sqrt(b))/b^(7/2) + 5/3*sqrt(b*x^2 + a*x)/b^3
Time = 0.29 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.53 \[ \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx=\frac {5 \, a \log \left ({\left | -2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} - a \right |}\right )}{2 \, b^{\frac {7}{2}}} + \frac {\sqrt {b x^{2} + a x}}{b^{3}} + \frac {2 \, {\left (9 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )}^{2} a^{2} b + 15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} a^{3} \sqrt {b} + 7 \, a^{4}\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a x}\right )} \sqrt {b} + a\right )}^{3} b^{\frac {7}{2}}} \]
5/2*a*log(abs(-2*(sqrt(b)*x - sqrt(b*x^2 + a*x))*sqrt(b) - a))/b^(7/2) + s qrt(b*x^2 + a*x)/b^3 + 2/3*(9*(sqrt(b)*x - sqrt(b*x^2 + a*x))^2*a^2*b + 15 *(sqrt(b)*x - sqrt(b*x^2 + a*x))*a^3*sqrt(b) + 7*a^4)/(((sqrt(b)*x - sqrt( b*x^2 + a*x))*sqrt(b) + a)^3*b^(7/2))
Timed out. \[ \int \frac {x^5}{\left (a x+b x^2\right )^{5/2}} \, dx=\int \frac {x^5}{{\left (b\,x^2+a\,x\right )}^{5/2}} \,d x \]